# Variation About the Mean Part B: Unfair Allocations (25 minutes)

In This Part: Fair and Unfair Allocations
The average for a set of data corresponds to the equal-shares allocation or fair allocation of the data. For example, suppose that each of 9 people has several dollars and altogether they have \$45. The mean of \$5 represents the number of dollars each of the 9 people would get if they combined all their money and then redistributed it fairly (i.e., equally).

As seen in Problem A3, the fair allocation of 45 coins into 9 stacks is to place 5 coins in each stack, as follows: Here is a second allocation of the 45 coins, which is almost fair: The above allocation is almost fair because most stacks have 5 coins, and the others have close to 5. But, the following allocation of 45 coins doesn’t seem fair at all: Problem B1
Look at these five allocations: Rank the five allocations on their “fairness,” from most fair to least fair. Also, explain how you decided on the level of “unfairness” in an allocation. Although there are many different ways to think about “unfairness,” one way is to consider the number of exchanges that would have to take place in order to achieve a fair allocation.

In This Part: Measuring the Degree of Fairness
In Problem B1, it’s easy to identify the most and least fair allocations. It’s more difficult to decide the degree of fairness among the remaining three allocations. Note 3

We need a more objective way to measure how close an allocation is to the fair allocation. Here’s one method:
• Arrange the stacks in the allocation in increasing order.
• Move 1 coin from an above-average stack to a below-average stack. (Here, “average” refers to the mean.) This move will create a new allocation of the coins with the same mean.
• Continue moving coins, one at a time, from an above-average stack to a below-average stack until (if possible) each stack contains the same number of coins.
• One measure of the degree of fairness for an allocation is the number of coins you had to move. The smaller this number is, the closer the allocation is to the fair allocation.

Problem B2
Below are three allocations of 45 coins in 9 stacks. For each allocation, find the minimum number of moves required to change the allocation into a fair allocation (i.e., one with 5 coins in each stack).

Allocation A Allocation B Allocation C Problem B3
Based on the minimum number of moves, which of the allocations in Problem B2 is the most fair? Which is the least fair?

In This Part: Looking at Excesses and Deficits
We’ve seen how to judge the relative fairness of an allocation. A related question asks how you might determine the number of moves required to make an unfair allocation fair, without making the actual moves. See Note 4 Below.

Below is Allocation A from Problem B2, arranged in ascending order. We’ve already determined that only two moves are required to make this allocation fair. Let’s look more closely at why this is true: The notations below each stack indicate the number of coins that each stack is above or below the mean of 5. In other words:

a. Two stacks are above the mean. Each of these has 6 coins, an excess of 1 coin (+1) above the average. (These excesses are noted in the figure above.) The total excess of coins above the average is 2.
b. Two stacks are below the mean. Each of these has 4 coins, a deficit of 1 coin (-1) below the average. (These deficits are also noted in the figure above.) The total deficit of coins below the average is 2.
c. Five stacks are exactly average and have no excess or deficit (0).

Problem B4
Perform the calculations above to show that 20 moves are required to obtain a fair allocation for Allocation B, which is arranged in ascending order below: Problem B5
Perform the same calculations to find the number of moves required to obtain a fair allocation for Allocation C, which is shown in ascending order below: ### Notes

Note 3
In this part, you consider the fairness of different allocations: Which allocation is the most fair? Which allocation is the least fair?

These questions ask for more than a distinction between fair and unfair — they suggest that there are degrees of unfairness. Come up with your own method for measuring degrees of fairness. Part of the point of this lesson is to realize that statistical measures are invented by people.

If you are working in a group, discuss several methods for measuring fairness.

### Solutions

Problem B1
Selecting the most fair allocation should be fairly easy. Most people immediately pick Allocation A because it has 5 stacks of size 5, and the other 4 stacks are of size 4 or 6. It is the most similar to the fair allocation.

Selecting the least fair allocation is also relatively easy. Most people immediately pick Allocation B as the least fair. Its stacks are only of size 1 or 10, which is quite different from the fair allocation of 5 for each stack.

Otherwise, answers will vary depending on how people measure “fairness.” One method is to find the allocation where, on average, the stack sizes are closest to 5. This is accomplished by finding, in total, how far away the stacks are from 5. Using this criterion, Allocation A is the most fair, followed by Allocations D, C, E, and B.

Problem B2
• Allocation A requires 2 moves.
• Allocation B requires 20 moves.
• Allocation C requires 7 moves.

Problem B3
Allocation A is the most fair, and Allocation B is the least fair. The number of moves required to make an allocation fair is one way to measure how close to being fair the allocation is. That is, the more moves required to make an allocation fair, the more unfair the allocation is.

Problem B4 Four stacks are above the mean. Each of these has 10 coins, an excess of 5 coins (+5) above the average. The total excess of coins above the average is 20.
Five stacks are below the mean. Each of these has 1 coin, a deficit of 4 coins (-4) below the average. The total deficit of coins below the average is 20.
• No stacks are exactly average.

Problem B5 • Four stacks are above the mean. The total excess of coins above the average is 7 ([+1] + [+1] + [+2] + [+3]).
• Four stacks are below the mean. The total deficit of coins below the average is 7 ([-3] + [-2] + [-1] + [-1]).
• One stack is exactly average and has no excess or deficit (0).